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Calcium carbonate decomposes into calcium oxide and carbon dioxide as shown below. If 20.7 grams of CaCO3 decompose, what is the theoretical yield in grams of CaO?

If 6.81 grams of CaO are actually recovered, what is the percent yield?


CaCO3 --> CaO + CO2

**Your answer should be written as XX.X

Respuesta :

Answer:

Percentage yield =58.7 %

Explanation:

moles of calcium carbonate as a reactant = [tex]\frac{20.7 g}{100.086 g/mol} =0.2068 mole[/tex]

as per molar equation , 0.2068 moles of calcium carbonate produces 0.2068 moles of CaO .

amount of CaO = 0.2068 mole x (40.078+15.999) g/moles = 11.598 gram

percentage yield = actual yield /  theoritical yield =  6.81/ 11.598 = 0.587 or 58.7 %